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Electric power systems

S. Vasantharathna Department of Electrical and Electronics Engineering, Coimbatore Institute of Technology, Coimbatore, Tamil Nadu, India

Abstract

Power supply systems are widely interconnected across regions for economic reasons, to reduce the cost of electricity and improve the reliability of power supply. At the same time, with growing technological advancements impacting the power industry, power supply systems may also lead to insecurity. Bottlenecks in interconnected power systems that may lead to security problems might include excessive reactive power, faults, dynamic swings of connected loads, and generation capacity. Hence, to provide dynamic control of system voltage, frequency, and reactive power, and to design the generation capacity and decide upon the rating of protective devices, this chapter will be of significant use to avoid outages and increase the reliability of power systems.

Keywords

one line diagram

per unit quantities

bus admittance matrix

bus impedance matrix

load flow analysis

fault analysis

stability analysis

18.1. Introduction

Basic components of a power system are generators, transformers, transmission lines, and loads. The interconnection of the components discussed in section 18.2 in a power system is represented in a one-line diagram or single-line diagram. The advantage of a one-line diagram is its simplicity. One line in the single line diagram represents single phase or/and all three phases of the balanced system. Equivalent circuits of the components are replaced by their standard symbols and completion of the circuit neutral is omitted. Figure 18.1 shows the one line diagram of a typical power system network.

Figure 18.2 [1] shows different configurations of the single-phase transformers that are commonly used. But in a one-line diagram, it is always represented symbollically as in Figure 18.3 irrespective of whether it is a single-phase or a three-phase transformer. A generating station may have one or more generators and a pool of generating stations synchronized as represented by a single circle in a one-line diagram. The generator, load, transmission line, and circuit breakers are represented in a single line diagram as shown in Figure 18.3. Figure 18.4 shows a model of a single line diagram of the power system network that can be modeled using a simulation tool (Figures 18.5 to 18.7).

Figure 18.2 Winding representations of single-phase transformers.

Figure 18.3 Symbolic representation of components in a single-line diagram.

An impedance diagram [4] is derived from the one-line diagram, representing the equivalent circuits of power system components. If resistances, static loads, and transmission line capacitances are neglected, it is known as a reactance diagram. The impedance diagram and reactance diagram is much helpful in load flow studies, fault studies and stability analysis of a power system network. The location of circuit breakers are not required for load flow studies.

The following assumptions are made while drawing the impedance and reactance diagram.

1. A generator can be represented by a voltage source in series with an inductive reactance. The internal resistance of the generator is negligible compared to the reactance.

2. The loads are inductive.

3. The transformer core is ideal and may be represented by a reactance.

4. The transmission line is a medium length line and can be denoted by a T or π circuit.

Problem 18.1

Generating station 1 is connected to a load A and is transmitting power through a transmission line. The receiving end of the transmission line is connected to load B and three other generating stations. Draw the one line diagram, impedance diagram, and reactance diagram.

Solution

The single line diagram of the given power system network is shown in Figure 18.5. The small squares represent the location of circuit breakers. The vertical lines are the bus. The impedance diagram shown in Figure 18.6 is derived by drawing the respective equivalent circuits. The resistance (R), inductive reactance (X_I) and capacitive reactance (X_c) are calculated based on standard formulae. The reactance diagram shown in Figure 18.7 is drawn by removing resistances and capacitive reactances as their impact on performance analysis is negligible.

18.1.1. Per unit quantities

Per unit quantities, like percentage quantities, are fractional quantities of a reference quantity used to reduce the computational complexity. Per unit values are written with “pu” after the value. For power, voltage, current, and impedance, a per unit quantity may be obtained by dividing the respective base or reference of that quantity.

$p u quantity = \frac{Actual quantity}{Base quantity} .$ $p u quantity = \frac{Actual quantity}{Base quantity} .$

The pu representation of the quantities viz., complex power, voltage, current, and impedance, respectively are given as follows.

$S_{pu} = \frac{S}{S_{base}}, V_{pu} = \frac{V}{V_{base}}, I_{pu} = \frac{I}{I_{base}}, and Z_{pu} = \frac{Z}{Z_{base}} .$ $S_{pu} = \frac{S}{S_{base}}, V_{pu} = \frac{V}{V_{base}}, I_{pu} = \frac{I}{I_{base}}, and Z_{pu} = \frac{Z}{Z_{base}} .$

Only two base or reference quantities need to be independently defined, because voltage, current, impedance, and power are related. The base quantities for the other two can be derived therefrom. Since power and voltage are most often specified, they are usually chosen to define the independent base quantities.

If VA_base and V_base are the selected base quantities of power (complex, active, or reactive) and voltage, respectively, then,

$\begin{array}{l} Base current I_{base} = \frac{V_{base} I_{base}}{V_{base}} = \frac{V A_{base}}{V_{base}} \\ Base impedance Z_{base} = Z_{base} = \frac{V_{base}}{I_{base}} = \frac{V_{base}^{2}}{I_{base} V_{base}} = \frac{V_{base}^{2}}{V A_{base}} . \end{array}$ $\begin{array}{l} Base current I_{base} = \frac{V_{base} I_{base}}{V_{base}} = \frac{V A_{base}}{V_{base}} \\ Base impedance Z_{base} = Z_{base} = \frac{V_{base}}{I_{base}} = \frac{V_{base}^{2}}{I_{base} V_{base}} = \frac{V_{base}^{2}}{V A_{base}} . \end{array}$

In a power system, voltages and power are usually expressed in kilovolts (kV) and megavolt amperes (MVA), thus it is usual to select an MVA_base and a kV_base and to express them as,

$\begin{array}{l} Base current I_{base} = \frac{{MVA}_{base}}{{kV}_{base}} in kA \\ Base impedance Z_{base} = \frac{{kV}_{base}^{2}}{{MVA}_{base}} in Ω . \end{array}$ $\begin{array}{l} Base current I_{base} = \frac{{MVA}_{base}}{{kV}_{base}} in kA \\ Base impedance Z_{base} = \frac{{kV}_{base}^{2}}{{MVA}_{base}} in Ω . \end{array}$

In these expressions, all the quantities are single-phase quantities. In three-phase systems the line voltage and total power are usually used rather than the single-phase quantities. It is thus usual to express base quantities in terms of these.

If VA_3Φbase and V_LLbase are base three-phase power and line-to-line voltage, respectively, then

$\begin{array}{l} Base current I_{base} = \frac{{MVA}_{3 Φ base}}{\sqrt{3} {kV}_{LLbase}} in kA \\ Base impedance Z_{base} = \frac{{kV}_{LLbase}^{2}}{{MVA}_{3 Φ base}} in Ω . \end{array}$ $\begin{array}{l} Base current I_{base} = \frac{{MVA}_{3 Φ base}}{\sqrt{3} {kV}_{LLbase}} in kA \\ Base impedance Z_{base} = \frac{{kV}_{LLbase}^{2}}{{MVA}_{3 Φ base}} in Ω . \end{array}$

Problem 18.2

Given the actual and base quantities, express the following quantities in pu form.

Actual quantities are 20 A, 0.2 A, 50 V, 1000 V, and 2 Ω.

Base quantities are 10 A, 200 V, and 20 Ω.

$\begin{array}{l} I_{pu} = \frac{20}{10} = 2 pu. \\ I_{pu} = \frac{0 .2}{10} = 0 .02 pu. \\ V_{pu} = \frac{50}{200} = 0 .25 pu. \\ V_{pu} = \frac{1000}{200} = 5 pu. \\ Z_{pu} = \frac{2}{20} = 0 .1 pu. \end{array}$ $\begin{array}{l} I_{pu} = \frac{20}{10} = 2 pu. \\ I_{pu} = \frac{0 .2}{10} = 0 .02 pu. \\ V_{pu} = \frac{50}{200} = 0 .25 pu. \\ V_{pu} = \frac{1000}{200} = 5 pu. \\ Z_{pu} = \frac{2}{20} = 0 .1 pu. \end{array}$

Problem 18.3

In the circuit shown in Figure 18.8, consider the base quantities of voltage and impedance as V_b = 100 V; Z_b = 0.01 Ω. Find I_b, I_pu, V_pu, Z_pu, and I.

$\begin{array}{l} Z & = 0.01 + j 0.01 Ω \\ I_{b} & = V_{b} / Z_{b} = 100 / 0.01 = 1 0^{4} A \\ V_{pu} & = 100 / 100 = 1 pu \\ Z_{pu} & = 0.01 + j 0.1 / 0.01 = 1 + j 1 pu \\ I_{pu} & = 1 / 1 + j1 = 0.5 - 0.5 pu . \end{array}$ $\begin{array}{l} Z & = 0.01 + j 0.01 Ω \\ I_{b} & = V_{b} / Z_{b} = 100 / 0.01 = 1 0^{4} A \\ V_{pu} & = 100 / 100 = 1 pu \\ Z_{pu} & = 0.01 + j 0.1 / 0.01 = 1 + j 1 pu \\ I_{pu} & = 1 / 1 + j1 = 0.5 - 0.5 pu . \end{array}$

Problem 18.4

Choosing a base MVA of 50 and a base kV of 33, find the pu value of 10 Ω resistance.

$\begin{array}{l} Z_{b} & = 332 / 50 = 21.78 Ω \\ Z_{pu} & = 10 / 21.78 = 0.45914 pu . \end{array}$ $\begin{array}{l} Z_{b} & = 332 / 50 = 21.78 Ω \\ Z_{pu} & = 10 / 21.78 = 0.45914 pu . \end{array}$

Problem 18.5

A three-phase, 13 kV transmission line delivers 8 MVA load. The per phase impedance of the line is 0.01 + j 0.05 pu. What is the voltage drop across the line, when it is referred to a 13 kV, 8 MVA_base?

The given base quantities yield,

$\begin{array}{l} Base kVA = 8000 = 1 pu \\ Base kV = 13 = 1 pu. \end{array}$ $\begin{array}{l} Base kVA = 8000 = 1 pu \\ Base kV = 13 = 1 pu. \end{array}$

Then the other base quantities are,

$\begin{array}{l} Base current = 8000 / 13 \\ Base current = \frac{8000}{13 \sqrt{3}} = 355.292 A \\ Base impedance = \frac{13000}{355.292} = 36.59 Ω \\ Impedance = 36 .59 (0 .01 + j 0 .05) = 0 .3659 + j 1 .8295 Ω \\ Voltage drop = 355 .292 (0 .3659 + j 1 .8295) = 130 .001 + j 650 .01 = 662 .88 V. \end{array}$ $\begin{array}{l} Base current = 8000 / 13 \\ Base current = \frac{8000}{13 \sqrt{3}} = 355.292 A \\ Base impedance = \frac{13000}{355.292} = 36.59 Ω \\ Impedance = 36 .59 (0 .01 + j 0 .05) = 0 .3659 + j 1 .8295 Ω \\ Voltage drop = 355 .292 (0 .3659 + j 1 .8295) = 130 .001 + j 650 .01 = 662 .88 V. \end{array}$

Conversions from one base to another: normally the per unit value is defined to its own rating. In a power system network, different components can have different ratings, and may be different from the system rating, therefore it is necessary to convert all quantities to a common base to perform numeric computations. Also if a new station is added/removed to/from a network, the reference quantities might get changed. Instead of recalculating the pu quantities based on new reference values, for all the systems, change of base is preferred. The conversion from one base to another in a system is as follows,

$Z_{pu} = Z_{old} \frac{{MVA}_{base new}}{{MVA}_{base old}} \frac{{kV}_{base old}^{2}}{{kV}_{base new}^{2}} .$ $Z_{pu} = Z_{old} \frac{{MVA}_{base new}}{{MVA}_{base old}} \frac{{kV}_{base old}^{2}}{{kV}_{base new}^{2}} .$

Problem 18.6

A 11 kV, 15 MVA generator has a reactance of 0.15 pu referred to its own ratings as base. The new bases chosen are 110 kV and 30 MVA. Calculate the new pu reactance:

$Z_{pu} = 0.15 \times \frac{30}{15} \times \frac{11^{2}}{110^{2}} = 0.003 pu .$ $Z_{pu} = 0.15 \times \frac{30}{15} \times \frac{11^{2}}{110^{2}} = 0.003 pu .$

Problem 18.7

Three generators are rated as follows. Draw the reactance diagram.

$\begin{array}{l} G1 = 100 MVA, 33 kV, X " = 10 % \\ G2 = 150 MVA, 32 kV, X " = 8 % \\ G3 = 110 MVA, 30 kV, X " = 12 % \\ Base = 200 MVA, 35 kV \\ X G1 pu = 0.1 \times \frac{200}{100} \times \frac{33^{2}}{35^{2}} = 0.1773 pu \\ X G2 pu = 0.08 \times \frac{200}{150} \times \frac{32^{2}}{35^{2}} = 0.0892 pu \\ X G3 pu = 0.12 \times \frac{200}{110} \times \frac{30^{2}}{35^{2}} = 0.1603 pu. \end{array}$ $\begin{array}{l} G1 = 100 MVA, 33 kV, X " = 10 % \\ G2 = 150 MVA, 32 kV, X " = 8 % \\ G3 = 110 MVA, 30 kV, X " = 12 % \\ Base = 200 MVA, 35 kV \\ X G1 pu = 0.1 \times \frac{200}{100} \times \frac{33^{2}}{35^{2}} = 0.1773 pu \\ X G2 pu = 0.08 \times \frac{200}{150} \times \frac{32^{2}}{35^{2}} = 0.0892 pu \\ X G3 pu = 0.12 \times \frac{200}{110} \times \frac{30^{2}}{35^{2}} = 0.1603 pu. \end{array}$

The reactance diagram is shown in Figure 18.9.

Figure 18.9 Reactance diagram for Problem 18.9.

Per unit representation of a transformer: Consider the equivalent circuit of the transformer shown in Figure 18.10.

Here, Z_p, leakage reactance on the primary side; Z_s, leakage reactance on the secondary side.

Transformation ratio = 1:a.

Choose VA_base and V_base on two sides of a transformer such that,

$\frac{V_{1 b}}{V_{2 b}} = 1 / a \frac{I_{1 b}}{I_{2 b}} = a Z_{1 b} = \frac{V_{1 b}}{I_{1 b}} Z_{2 b} = \frac{V_{2 b}}{I_{2 b}} .$ $\frac{V_{1 b}}{V_{2 b}} = 1 / a \frac{I_{1 b}}{I_{2 b}} = a Z_{1 b} = \frac{V_{1 b}}{I_{1 b}} Z_{2 b} = \frac{V_{2 b}}{I_{2 b}} .$

From Figure 18.10, it is written as V₂ = (V₁ – I₁Z_p)a – (I₂Z_s).

In pu form, V_2puV_2b = [V_1puV_1b – I_1puI_1bZ_ppuZ_1b]a – I_2puI_2bZ_spuZ_2b.

Divide by V_2b throughout using the base relation V_2pu = V_1pu – I_1pu Z_ppu – I_2pu Z_spu.

Using the relations

\frac{I_{1}}{I_{2}} = \frac{I_{1 b}}{I_{2 b}} = a, \frac{I_{1}}{I_{1 b}} = \frac{I_{2}}{I_{2 b}} \to I_{1 pu} = I_{2 pu} = I_{pu}

$\frac{I_{1}}{I_{2}} = \frac{I_{1 b}}{I_{2 b}} = a, \frac{I_{1}}{I_{1 b}} = \frac{I_{2}}{I_{2 b}} \to I_{1 pu} = I_{2 pu} = I_{pu}$

, it is rewritten as,

$V_{2pu} = V_{1pu} - I_{pu} Z_{pu} .$ $V_{2pu} = V_{1pu} - I_{pu} Z_{pu} .$

Where, Z_pu = Z_ppu + Z_spu.

Even from the primary or secondary side pu Z can be calculated.

On the primary side,

$\begin{array}{l} Z_{1} = Z_{p} + \frac{Z_{s}}{a^{2}} \\ Z_{1 pu} = \frac{Z_{1}}{Z_{1 b}} = \frac{Z_{p}}{Z_{1 b}} + \frac{Z_{s} / a^{2}}{Z_{1 b}} = \frac{Z_{p}}{Z_{1 b}} + \frac{Z_{s}}{Z_{1 b} a^{2}} \\ ∴ Z_{1pu} = Z_{ppu} + Z_{spu} = Z_{pu} . \end{array}$ $\begin{array}{l} Z_{1} = Z_{p} + \frac{Z_{s}}{a^{2}} \\ Z_{1 pu} = \frac{Z_{1}}{Z_{1 b}} = \frac{Z_{p}}{Z_{1 b}} + \frac{Z_{s} / a^{2}}{Z_{1 b}} = \frac{Z_{p}}{Z_{1 b}} + \frac{Z_{s}}{Z_{1 b} a^{2}} \\ ∴ Z_{1pu} = Z_{ppu} + Z_{spu} = Z_{pu} . \end{array}$

On the secondary side:

$\begin{array}{l} Z_{2} = Z_{s} + a^{2} Z_{p} \\ Z_{2 pu} = \frac{Z_{2}}{Z_{2 b}} = \frac{Z_{s}}{Z_{2 b}} + a^{2} \frac{Z_{p}}{Z_{2 b}} \\ ∴ Z_{2pu} = Z_{spu} + Z_{ppu} = Z_{pu} . \end{array}$ $\begin{array}{l} Z_{2} = Z_{s} + a^{2} Z_{p} \\ Z_{2 pu} = \frac{Z_{2}}{Z_{2 b}} = \frac{Z_{s}}{Z_{2 b}} + a^{2} \frac{Z_{p}}{Z_{2 b}} \\ ∴ Z_{2pu} = Z_{spu} + Z_{ppu} = Z_{pu} . \end{array}$

Therefore, pu impedance of a transformer is the same whether computed from the primary or secondary side, as long as the voltage bases on two sides are the ratio of transformation.

Problem 18.8

A generating station is supplying power to a distant village 50 km away. The transmission is done on 110 kV transmission line. The generator is rated at 400 MVA giving the output at 11 kV and has a subtransient reactance of 20%. The load consists of motors running at 11 kV and rated for 60, 80, and 100 MVA. The subtransient reactance of motors is 18%. The transformer at the generating station is rated for 300 MVA with a leakage reactance of 10% and a voltage rating of 11/110 kV. The transformer at the village is rated for 250 MVA with a leakage reactance of 12% and a voltage rating of 110/11 kV. The reactance of the line is 0.1 Ω/km. Draw the reactance diagram of the system.

The one line diagram of the descripted power system network is shown in Figure 18.11. G1: 400 MVA, 11 kV selected as base.

Base value in the transmission line = 11 × (110/11) = 110 kV.

Base value in the motor = 110 × (11/110) = 11 kV.

$Z_{pu} = Z_{old} \frac{M V A_{base new}}{M V A_{base old}} \frac{k V_{base old}^{2}}{k V_{base new}^{2}} .$ $Z_{pu} = Z_{old} \frac{M V A_{base new}}{M V A_{base old}} \frac{k V_{base old}^{2}}{k V_{base new}^{2}} .$

Reactance of a transformer based on its own rating, converted to the common base quantity is,

$\begin{array}{l} X T1 pu = 0.1 \times \frac{400}{300} \times \frac{11^{2}}{11^{2}} = 0.1333 pu \\ X T2 pu = 0.12 \times \frac{400}{250} \times \frac{110^{2}}{110^{2}} = 0.1920 pu \\ X TL pu = 0.1 \times 50 \times \frac{400}{110^{2}} = 0.1652 pu \\ X m1 pu = 0.18 \times \frac{400}{60} \times \frac{11^{2}}{11^{2}} = 1.2 pu \\ X m2 pu = 0.18 \times \frac{400}{80} \times \frac{11^{2}}{11^{2}} = 0.9 pu \\ X m1 pu = 0.18 \times \frac{400}{100} \times \frac{11^{2}}{11^{2}} = 0.72 pu \end{array}$ $\begin{array}{l} X T1 pu = 0.1 \times \frac{400}{300} \times \frac{11^{2}}{11^{2}} = 0.1333 pu \\ X T2 pu = 0.12 \times \frac{400}{250} \times \frac{110^{2}}{110^{2}} = 0.1920 pu \\ X TL pu = 0.1 \times 50 \times \frac{400}{110^{2}} = 0.1652 pu \\ X m1 pu = 0.18 \times \frac{400}{60} \times \frac{11^{2}}{11^{2}} = 1.2 pu \\ X m2 pu = 0.18 \times \frac{400}{80} \times \frac{11^{2}}{11^{2}} = 0.9 pu \\ X m1 pu = 0.18 \times \frac{400}{100} \times \frac{11^{2}}{11^{2}} = 0.72 pu \end{array}$

The reactance diagram is shown in Figure 18.12.

18.2. Phases of power system engineering

The power system network includes generation, transmission, distribution, and utilization systems as shown in Figure 18.13. In the generation subsystem, the power plant produces the electricity. The transmission subsystem transmits the electricity to the load centers. The distribution subsystem distributes the power to the customers. The utilization system is concerned with the different uses of electrical energy.

The grid is an electrical network that connects a variety of electric generators in different generating stations located remotely to the users of electric power. In high voltage transmission lines step-up transformers are used to reduce transmission power loss. Substations convert power to higher voltages before transmission and to lower voltages suitable for appliances after transmission [5].

18.3. Interconnected systems

Power generating stations are interconnected to improve the reliability of the supply, to reduce the reserve capacity required, and to improve the load factor, diversity factor, and the overall efficiency. More efficient plants are operated as base load plants and less efficient plants are operated as peak load plants, such that the generation cost and capital cost per kW are reduced. In the deregulated environment, interconnection of generating stations is being supported to enable the exchange of power among regions or countries and transport cheaper energy over long distances to the load centers. For this exchange of power through the interconnected grid, it must be ensured that all the generators run not only at the same frequency, but also at the same phase. A large failure in one part of the grid, if not quickly compensated, may lead to cascaded failures. Load dispatch centers facilitate communication between generating stations to maintain a stable grid. HVDC or variable frequency transformers can be used to connect two alternating current interconnection networks, which are not synchronized with each other, without the need to synchronize the wider area. However, there are technical and economical limitations in the interconnections, if the power has to be transmitted over extremely long distances.

18.3.1. Microgrids [2]

The need for a sustainable and reliable power supply and the availability of renewable energy sources lead to the development of the microgrid. Combinations of different but complementary energy generation systems based on renewable or mixed energy (renewable energy with a backup biofuel/biodiesel generator) are known as renewable energy hybrid systems. The grid formed by this system is known as a microgrid due to its size compared to the main grid thereby reducing the transmission and distribution losses.

A typical microgrid configuration is shown in Figure 18.14. It consists of electrical/heat loads and microsources connected through a low-voltage distribution network. The microgrid is provided with systems to implement the control, metering, and protection functions during standalone and grid-connected modes of operation. In Figure 18.14, the microgrid consists of radial feeders to supply the electrical and heat loads, which can also be categorized under priority (that requires uninterrupted power supply) and nonpriority loads. Microsources and storage devices are connected to feeders through microsource controllers (MCs). The microgrid is coupled with the main medium voltage (MV) utility grid through the point of common coupling (PCC). A circuit breaker is operated to connect and disconnect the entire microgrid from the main grid, as per the selected mode of operation, such as, (1) grid connected and (2) standalone.

Figure 18.14 Typical microgrid configuration.

In grid-connected mode, the microgrid remains connected to the main grid either totally or partially, and imports or exports power from or to the main grid. The operation and management of the microgrid in different modes is controlled and coordinated through a local MC and the central controller (CC).

1. MC – The main function of the MC is to independently control the power flow and load end voltage profile of the microsource in response to any disturbance and load changes. The MC also participates in economic generation scheduling, load tracking/management and demand-side management by controlling the storage devices. The built-in control features of the MCs are:

a. Active and reactive power control.

b. Voltage control.

c. Storage requirement for fast load tracking.

d. Load sharing through power–frequency (P–f) control.

2. CC – Two main functional modules of CC are the energy management module (EMM) and the protection coordination module (PCM). The objectives of the CC are,

a. to maintain a specified voltage and frequency at the load end through P–f and voltage control,

b. to ensure energy optimization for the microgrid,

c. to perform protection coordination, and

d. to provide the power dispatch and voltage set points for all the MCs.

- EMM provides the set points for active and reactive power output, voltage, and frequency to each MC.

- PCM responds to the microgrid and main grid faults and loss of grid scenarios in a way so as to ensure correct protection coordination of the microgrid. It also adapts to the change in fault current levels during changeover from grid-connected to standalone mode.

The functions of CC in grid-connected mode are as follows.

1. Collects information from the microsources and loads connected to the microgrid and monitors system diagnostics.

2. Performs state estimation and security assessment evaluation, economic generation scheduling, active and reactive power control of the microsources, and demand-side management functions, by using collected information.

3. Ensures synchronized operation with the main grid maintaining the power exchange at prior contract points.

The functions of the CC in standalone mode are as follows.

1. Performs active and reactive power control of the microsources in order to maintain stable voltage and frequency at load ends.

2. Adopts load interruption/shedding strategies using demand-side management with storage device support to maintain power balance and bus voltage.

3. Initiates a local black start to ensure improved reliability and continuity of service.

4. Switches over the microgrid to grid-connected mode after main grid supply is restored, without hampering the stability of either grid.

Supervisory control design philosophy includes optimal dispatch control to provide P&Q set point with a slow time constant of 5–10 min and timeline control to manage voltage and power at the point of interconnection and to provide control set points for local control with a faster time constant of 10–100 ms. For the load flow analysis, bus admittance matrix need to be formulated. For fault studies, Bus impedance matrix need to be formulated.

18.3.2. Bus admittance matrix [5]

The bus admittance matrix is formulated to load flow analysis.

Generator power: S_Gi = P_Gi + jQ_Gi and load power: S_Li = P_Li + jQ_Li.

Complex power: S_i = S_Gi – S_Li = (P_Gi – P_Li) + j(Q_Gi – Q_Li) = P_Li + jQ_Li.

Consider the one-line diagram and its equivalent impedance diagram of a four-bus system shown in Figure 18.15. The circuit with common reference at ground potential is redrawn as shown in Figure 18.16.

Figure 18.15 (a) One line diagram of a four-bus system. (b) Impedance diagram of a four-bus system.

Applying KCL to four nodes:

$\begin{array}{l} I_{1} = V_{1} y_{10} + (V_{1} - V_{2}) y_{12} + (V_{1} - V_{3}) y_{12} + (V_{1} - V_{4}) y_{14} \\ I_{2} = V_{2} y_{20} + (V_{2} - V_{1}) y_{12} + (V_{2} - V_{3}) y_{23} \\ I_{3} = V_{3} y_{20} + (V_{3} - V_{1}) y_{13} + (V_{3} - V_{2}) y_{23} + (V_{3} - V_{4}) y_{34} \\ I_{4} = V_{4} y_{40} + (V_{4} - V_{1}) y_{14} + (V_{4} - V_{3}) y_{34} . \end{array}$ $\begin{array}{l} I_{1} = V_{1} y_{10} + (V_{1} - V_{2}) y_{12} + (V_{1} - V_{3}) y_{12} + (V_{1} - V_{4}) y_{14} \\ I_{2} = V_{2} y_{20} + (V_{2} - V_{1}) y_{12} + (V_{2} - V_{3}) y_{23} \\ I_{3} = V_{3} y_{20} + (V_{3} - V_{1}) y_{13} + (V_{3} - V_{2}) y_{23} + (V_{3} - V_{4}) y_{34} \\ I_{4} = V_{4} y_{40} + (V_{4} - V_{1}) y_{14} + (V_{4} - V_{3}) y_{34} . \end{array}$

In matrix form, it is written as,

$\begin{array}{l} [\begin{array}{c} I_{1} \\ I_{2} \\ I_{3} \\ I_{4} \end{array}] = [\begin{array}{c} y_{10} + y_{12} + y_{13} + y_{14} & - y_{12} & - y_{13} & - y_{14} \\ - y_{12} & y_{20} + y_{12} + y_{23} & - y_{23} & 0 \\ - y_{13} & - y_{23} & y_{30} + y_{13} + y_{23} + y_{34} & - y_{34} \\ - y_{14} & 0 & - y_{34} & y_{40} + y_{14} + y_{34} \end{array}] [\begin{array}{c} V_{1} \\ V_{2} \\ V_{3} \\ V_{4} \end{array}] \\ [\begin{array}{c} I_{1} \\ I_{2} \\ I_{3} \\ I_{4} \end{array}] = [\begin{array}{c} y_{11} & y_{12} & y_{13} & y_{14} \\ y_{21} & y_{22} & y_{32} & y_{42} \\ y_{31} & y_{32} & y_{33} & y_{34} \\ y_{41} & y_{42} & y_{43} & y_{44} \end{array}] [\begin{array}{c} V_{1} \\ V_{2} \\ V_{3} \\ V_{4} \end{array}] \to I_{bus} = Y_{bus} \cdot V_{bus} \end{array}$ $\begin{array}{l} [\begin{array}{c} I_{1} \\ I_{2} \\ I_{3} \\ I_{4} \end{array}] = [\begin{array}{c} y_{10} + y_{12} + y_{13} + y_{14} & - y_{12} & - y_{13} & - y_{14} \\ - y_{12} & y_{20} + y_{12} + y_{23} & - y_{23} & 0 \\ - y_{13} & - y_{23} & y_{30} + y_{13} + y_{23} + y_{34} & - y_{34} \\ - y_{14} & 0 & - y_{34} & y_{40} + y_{14} + y_{34} \end{array}] [\begin{array}{c} V_{1} \\ V_{2} \\ V_{3} \\ V_{4} \end{array}] \\ [\begin{array}{c} I_{1} \\ I_{2} \\ I_{3} \\ I_{4} \end{array}] = [\begin{array}{c} y_{11} & y_{12} & y_{13} & y_{14} \\ y_{21} & y_{22} & y_{32} & y_{42} \\ y_{31} & y_{32} & y_{33} & y_{34} \\ y_{41} & y_{42} & y_{43} & y_{44} \end{array}] [\begin{array}{c} V_{1} \\ V_{2} \\ V_{3} \\ V_{4} \end{array}] \to I_{bus} = Y_{bus} \cdot V_{bus} \end{array}$

where, y_ii, self-admittance or driving point – summation of all the admittances connected to the bus, y_ip, off diagonal – mutual or transfer admittance – negative of the connected admittance.

$V_{bus} = Z_{bus} \cdot I_{bus} → Z_{bus} = {Y_{bus}}^{- 1}$ $V_{bus} = Z_{bus} \cdot I_{bus} → Z_{bus} = {Y_{bus}}^{- 1}$

Problem 18.9

Formulate Y_bus for the four-bus system shown in Figure 18.17. Consider bus 4 as a reference bus.

The shunt admittance at the buses is negligible. The line impedances are as shown here.

Line bus to bus	1–2	2–3	3–4	1–4
R in pu	0.025	0.02	0.05	0.04
X in pu	0.1	0.08	0.2	0.16

Solution

$G_{bus} = \frac{R}{R^{2} + X^{2}} B = \frac{- X}{R^{2} + X^{2}} Y = G + j B$ $G_{bus} = \frac{R}{R^{2} + X^{2}} B = \frac{- X}{R^{2} + X^{2}} Y = G + j B$

Line bus to bus	1–2	2–3	3–4	1–4
G in pu	2.35	2.94	1.176	1.47
B in pu	–9.41	–11.76	–4.706	–5.88

$\begin{array}{l} Y_{11} = Y_{12} + Y_{14} Y_{22} = Y_{12} + Y_{23} Y_{33} = Y_{23} + Y_{34} \\ Y_{12} = Y_{21} = - Y_{12} Y_{23} = Y_{32} = - Y_{23} \end{array}$ $\begin{array}{l} Y_{11} = Y_{12} + Y_{14} Y_{22} = Y_{12} + Y_{23} Y_{33} = Y_{23} + Y_{34} \\ Y_{12} = Y_{21} = - Y_{12} Y_{23} = Y_{32} = - Y_{23} \end{array}$

Y₁₃ = Y₃₁ = –Y₁₃ = 0 as there is no connection.

$Y_{bus} = [\begin{array}{c} 3.82 - j 15.29 & - 2 .35 + j 9.41 & 0 \\ - 2.35 + j 9 .41 & 5 .29 - j 21.17 & - 2.94 + j 11.76 \\ 0 & - 2.94 + j 11.76 & 4 .116 - j 16.466 \end{array}] .$ $Y_{bus} = [\begin{array}{c} 3.82 - j 15.29 & - 2 .35 + j 9.41 & 0 \\ - 2.35 + j 9 .41 & 5 .29 - j 21.17 & - 2.94 + j 11.76 \\ 0 & - 2.94 + j 11.76 & 4 .116 - j 16.466 \end{array}] .$

Problem 18.10

The parameters of a four-bus system are as shown below. Draw the network and find the bus admittance matrix.

Bus code	Line impedance (pu)	Charging admittance (pu)
1–2	0.2 + j0.8	j0.02
2–3	0.3 + j0.9	j0.03
2–4	0.25 + j1	j0.04
3–4	0.2 + j0.8	j0.02
1–3	0.1 + j0.4	j0.01

Solution

The Figure 18.18 shows the network defined.

$Y_{12} = \frac{1}{Z_{12}} = \frac{1}{0.2 + j 0.8} = 0.294 - j 1.176 pu$ $Y_{12} = \frac{1}{Z_{12}} = \frac{1}{0.2 + j 0.8} = 0.294 - j 1.176 pu$

Figure 18.18 Impedance diagram of the four-bus system defined in Problem 18.11.

Shunt admittance: y₁₀ = j0.03, y₂₀ = j0.09, y₃₀ = j0.06, y₄₀ = j0.06 pu.

$Y_{bus} = [\begin{array}{c} 0.882 - j 3.498 & - 0.294 + j 1.176 & - 0.588 + j 2.352 & 0 \\ - 0.294 + j 1.176 & 0.862 - j 3.026 & 0.333 + j 1 & - 0.235 + j 0.94 \\ - 0.588 + j 2.352 & - 0.333 + j 1 & 1 .215 - j 4.468 & - 0.294 + j 1.1.76 \\ 0 & - 0.235 + j 0.94 & - 0.294 + j 1.176 & 0 .529 - j 2.056 \end{array}] .$ $Y_{bus} = [\begin{array}{c} 0.882 - j 3.498 & - 0.294 + j 1.176 & - 0.588 + j 2.352 & 0 \\ - 0.294 + j 1.176 & 0.862 - j 3.026 & 0.333 + j 1 & - 0.235 + j 0.94 \\ - 0.588 + j 2.352 & - 0.333 + j 1 & 1 .215 - j 4.468 & - 0.294 + j 1.1.76 \\ 0 & - 0.235 + j 0.94 & - 0.294 + j 1.176 & 0 .529 - j 2.056 \end{array}] .$

18.3.3. Bus impedance matrix [6]

Fault studies use an impedance matrix. Islanding can be detected by monitoring impedance changes. Consider the N-port network shown in Figure 18.19. Z_ij represents impedance between bus i and bus j.

$[\begin{array}{c} V_{1} \\ V_{2} \\ V_{i} \\ V_{n} \end{array}] = [\begin{array}{c} z_{11} & z_{12} & \dots & z_{1 n} \\ z_{21} & z_{22} & \dots & z_{2 n} \\ z_{i 1} & z_{i 2} & \dots & z_{i n} \\ z_{n 1} & z_{n 2} & \dots & z_{n n} \end{array}] [\begin{array}{c} I_{1} \\ I_{2} \\ I_{i} \\ I_{n} \end{array}] \to V_{bus} = Z_{bus} \cdot I_{bus} and Z_{i k} = Z_{k} diagonally symmetric .$ $[\begin{array}{c} V_{1} \\ V_{2} \\ V_{i} \\ V_{n} \end{array}] = [\begin{array}{c} z_{11} & z_{12} & \dots & z_{1 n} \\ z_{21} & z_{22} & \dots & z_{2 n} \\ z_{i 1} & z_{i 2} & \dots & z_{i n} \\ z_{n 1} & z_{n 2} & \dots & z_{n n} \end{array}] [\begin{array}{c} I_{1} \\ I_{2} \\ I_{i} \\ I_{n} \end{array}] \to V_{bus} = Z_{bus} \cdot I_{bus} and Z_{i k} = Z_{k} diagonally symmetric .$

A step-by-step algorithm to build the Z_bus has the benefit of avoiding recomputation of Z_bus, even if there is an addition/removal of a bus in a network. The four modifications that are possible with the existing Z_bus are as follows.

1. Adding self-impedance Z_s from a new bus to reference.

2. Adding Z_s from a new bus to an old bus.

3. Adding Z_s from an old bus to reference.

4. Adding Z_s between old buses.

Type 1 modification – Addition of a tree branch Z_s from a new bus q to reference as shown in Figure 18.20:

$\begin{array}{l} [\begin{array}{c} V_{1} \\ V_{2} \\ V_{i} \\ \begin{array}{l} V_{n} \\ V_{q} \end{array} \end{array}] = [\begin{array}{c} z_{11} & z_{12} & \dots & z_{1 n} \\ z_{21} & z_{22} & \dots & z_{2 n} \\ z_{i 1} & z_{i 2} & \dots & z_{i n} \\ z_{n 1} & z_{n 2} & \dots & z_{n n} \\ 0 & 0 & 0 & z_{s} \end{array}] [\begin{array}{c} I_{1} \\ I_{2} \\ I_{i} \\ \begin{array}{l} I_{n} \\ I_{q} \end{array} \end{array}] \\ Z_{i q} = Z_{q i} = 0, Z_{q q} = Z_{s} \\ {[z_{bus}]}_{new} = [\begin{array}{c} {[z_{bus}]}_{old} & 0 \\ 0 & z_{s} \end{array}] \end{array}$ $\begin{array}{l} [\begin{array}{c} V_{1} \\ V_{2} \\ V_{i} \\ \begin{array}{l} V_{n} \\ V_{q} \end{array} \end{array}] = [\begin{array}{c} z_{11} & z_{12} & \dots & z_{1 n} \\ z_{21} & z_{22} & \dots & z_{2 n} \\ z_{i 1} & z_{i 2} & \dots & z_{i n} \\ z_{n 1} & z_{n 2} & \dots & z_{n n} \\ 0 & 0 & 0 & z_{s} \end{array}] [\begin{array}{c} I_{1} \\ I_{2} \\ I_{i} \\ \begin{array}{l} I_{n} \\ I_{q} \end{array} \end{array}] \\ Z_{i q} = Z_{q i} = 0, Z_{q q} = Z_{s} \\ {[z_{bus}]}_{new} = [\begin{array}{c} {[z_{bus}]}_{old} & 0 \\ 0 & z_{s} \end{array}] \end{array}$

Type 2 modification – Addition of a tree branch Z_s from a new bus q to an old bus k as shown in Figure 18.21:

$\begin{array}{l} V_{q} & = Z_{s} I_{q} + V_{k} \\ = Z_{s} I_{q} + Z_{k1} I_{1} + Z_{k 2} I_{2} + \dots + Z_{kk} (I_{k} + I_{q}) + \dots + Z_{k n} I_{n} \\ = Z_{k 1} I_{1} + Z_{k 2} I_{2} + \dots + Z_{kk} I_{k} + \dots + Z_{k n} I_{n} + (Z_{kk} + Z_{s}) I_{q} . \end{array}$ $\begin{array}{l} V_{q} & = Z_{s} I_{q} + V_{k} \\ = Z_{s} I_{q} + Z_{k1} I_{1} + Z_{k 2} I_{2} + \dots + Z_{kk} (I_{k} + I_{q}) + \dots + Z_{k n} I_{n} \\ = Z_{k 1} I_{1} + Z_{k 2} I_{2} + \dots + Z_{kk} I_{k} + \dots + Z_{k n} I_{n} + (Z_{kk} + Z_{s}) I_{q} . \end{array}$

Similarly,

$\begin{array}{l} V_{1} & = Z_{11} I_{1} + Z_{12} I_{2} + Z_{13} I_{3} + \dots + Z_{1 k} (I_{k} + I_{q}) + \dots + Z_{1 n} I_{n} \\ = Z_{11} I_{1} + Z_{12} I_{2} + \dots + Z_{1 k} I_{k} + Z_{1 n} I_{n} + Z_{1 k} I_{q} . \end{array}$ $\begin{array}{l} V_{1} & = Z_{11} I_{1} + Z_{12} I_{2} + Z_{13} I_{3} + \dots + Z_{1 k} (I_{k} + I_{q}) + \dots + Z_{1 n} I_{n} \\ = Z_{11} I_{1} + Z_{12} I_{2} + \dots + Z_{1 k} I_{k} + Z_{1 n} I_{n} + Z_{1 k} I_{q} . \end{array}$

Similarly,

V₂, V₃, … V_n

can be expanded and in matrix form.

Type 3 modification – Addition of a link Z_s between an old bus k and reference, has been shown in Figure 18.22. This is an extension of type 2 modification. If node q is eliminated, it results in type 3 modification.

If I_q is eliminated the last row and last column are eliminated resulting in,

${[z_{bus}]}_{new} = {[z_{bus}]}_{old} - \frac{1}{z_{kk} + z_{s}} [\begin{array}{c} z_{1k} \\ z_{2k} \\ ⋮ \\ z_{n k} \end{array}] [\begin{array}{c} z_{k1} & z_{k2} & \dots & z_{k n} \end{array}]$ ${[z_{bus}]}_{new} = {[z_{bus}]}_{old} - \frac{1}{z_{kk} + z_{s}} [\begin{array}{c} z_{1k} \\ z_{2k} \\ ⋮ \\ z_{n k} \end{array}] [\begin{array}{c} z_{k1} & z_{k2} & \dots & z_{k n} \end{array}]$

Type 4 modification – Addition of a link Z_s between two old buses i and k are shown in Figure 18.23:

$\begin{array}{l} V_{1} = Z_{11} I_{1} + Z_{12} I_{2} + Z_{1i} (I_{i} + I_{q}) + Z_{1j} I_{j} + Z_{1k} (I_{k} - I_{q}) + \dots Z_{1n} I_{n} \\ V_{1} = Z_{11} I_{1} + Z_{12} I_{2} + \dots + Z_{1n} I_{n} + I_{q} (Z_{1i} - Z_{1k}) \\ V_{k} = Z_{s} I_{q} + V_{i} . \end{array}$ $\begin{array}{l} V_{1} = Z_{11} I_{1} + Z_{12} I_{2} + Z_{1i} (I_{i} + I_{q}) + Z_{1j} I_{j} + Z_{1k} (I_{k} - I_{q}) + \dots Z_{1n} I_{n} \\ V_{1} = Z_{11} I_{1} + Z_{12} I_{2} + \dots + Z_{1n} I_{n} + I_{q} (Z_{1i} - Z_{1k}) \\ V_{k} = Z_{s} I_{q} + V_{i} . \end{array}$

Expanding the equation for V_k:

$\begin{array}{l} Z_{k1} I_{1} + Z_{k2} I_{2} + \dots + Z_{k i} (I_{i} + I_{q}) + Z_{k j} I_{j} + Z_{kk} (I_{k} - I_{q}) + \dots \\ = Z_{s} I_{q} + Z_{i 1} I_{1} + Z_{i 2} I_{2} + \dots + Z_{i i} (I_{i} + I_{q}) + Z_{i j} I_{j} + Z_{i k} (I_{k} - I_{q}) + \dots \end{array}$ $\begin{array}{l} Z_{k1} I_{1} + Z_{k2} I_{2} + \dots + Z_{k i} (I_{i} + I_{q}) + Z_{k j} I_{j} + Z_{kk} (I_{k} - I_{q}) + \dots \\ = Z_{s} I_{q} + Z_{i 1} I_{1} + Z_{i 2} I_{2} + \dots + Z_{i i} (I_{i} + I_{q}) + Z_{i j} I_{j} + Z_{i k} (I_{k} - I_{q}) + \dots \end{array}$

Similarly for all the buses rearranging,

$0 = (Z_{i 1} - Z_{k1}) I_{1} + \dots + (Z_{i i} - Z_{k i}) I_{i} + (Z_{i j} - Z_{k j}) I_{j} + (Z_{i k} - Z_{kk}) I_{k} + \dots + (Z_{s} + Z_{i i} - Z_{i k} - Z_{k i} + Z_{kk}) I_{q}$ $0 = (Z_{i 1} - Z_{k1}) I_{1} + \dots + (Z_{i i} - Z_{k i}) I_{i} + (Z_{i j} - Z_{k j}) I_{j} + (Z_{i k} - Z_{kk}) I_{k} + \dots + (Z_{s} + Z_{i i} - Z_{i k} - Z_{k i} + Z_{kk}) I_{q}$

Eliminating I_q,

${[z_{bus}]}_{new} = {[z_{bus}]}_{old} - \frac{1}{z_{s} + z_{i i} + z_{kk} - 2 z_{i k}} [\begin{array}{c} z_{1 i} - z_{1k} \\ z_{2 i} - z_{2k} \\ ⋮ \\ z_{n i} - z_{nk} \end{array}] [\begin{array}{c} (z_{i 1} - z_{k1}) & \dots & (z_{i n} - z_{kn}) \end{array}]$ ${[z_{bus}]}_{new} = {[z_{bus}]}_{old} - \frac{1}{z_{s} + z_{i i} + z_{kk} - 2 z_{i k}} [\begin{array}{c} z_{1 i} - z_{1k} \\ z_{2 i} - z_{2k} \\ ⋮ \\ z_{n i} - z_{nk} \end{array}] [\begin{array}{c} (z_{i 1} - z_{k1}) & \dots & (z_{i n} - z_{kn}) \end{array}]$

Problem 18.11

Figure 18.24 shows a four-bus system. Treating bus four as the reference bus, obtain Z_bus.

Step I. Bus four is considered as the reference bus: add new bus one to reference: type 1 modification [z_bus] = [1].

Step II. Connect new bus two to reference: type 1 modification,

$Z_{bus} = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$ $Z_{bus} = [\begin{array}{c} 1 & 0 \\ 0 & 1 \end{array}]$

Step III. Connect bus three to reference: type 1 modification,

$Z_{bus} = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$ $Z_{bus} = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}]$

Step IV. Connect bus one to bus two: type 4 modification,

$Z_{bus} = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] - \frac{1}{1 + 1 + 1} [\begin{array}{c} 1 \\ - 1 \\ 0 \end{array}] [\begin{array}{c} 1 & - 1 & 0 \end{array}] \to Z_{bus} = [\begin{array}{c} \frac{2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & 1 \end{array}] .$ $Z_{bus} = [\begin{array}{c} 1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{array}] - \frac{1}{1 + 1 + 1} [\begin{array}{c} 1 \\ - 1 \\ 0 \end{array}] [\begin{array}{c} 1 & - 1 & 0 \end{array}] \to Z_{bus} = [\begin{array}{c} \frac{2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & 1 \end{array}] .$

Step V. Connect bus one to bus three: type 4 modification,

$Z_{bus} = [\begin{array}{c} \frac{2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & 1 \end{array}] - \frac{1}{1 + \frac{2}{3} + 1} [\begin{array}{c} \frac{2}{3} \\ \frac{1}{3} \\ - 1 \end{array}] [\begin{array}{c} \frac{2}{3} & \frac{1}{3} & - 1 \end{array}] = [\begin{array}{c} \frac{1}{2} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{5}{3} & \frac{1}{8} \\ \frac{1}{4} & \frac{1}{8} & \frac{5}{8} \end{array}] .$ $Z_{bus} = [\begin{array}{c} \frac{2}{3} & \frac{1}{3} & 0 \\ \frac{1}{3} & \frac{2}{3} & 0 \\ 0 & 0 & 1 \end{array}] - \frac{1}{1 + \frac{2}{3} + 1} [\begin{array}{c} \frac{2}{3} \\ \frac{1}{3} \\ - 1 \end{array}] [\begin{array}{c} \frac{2}{3} & \frac{1}{3} & - 1 \end{array}] = [\begin{array}{c} \frac{1}{2} & \frac{1}{4} & \frac{1}{4} \\ \frac{1}{4} & \frac{5}{3} & \frac{1}{8} \\ \frac{1}{4} & \frac{1}{8} & \frac{5}{8} \end{array}] .$

18.4. Fault analysis

Fault analysis in power systems is required to design the rating of protective devices and to study the stability of a system. Fault studies need to be routinely performed, as the power system network is dynamic with the addition or removal of generators, transmission lines, and loads.

Faults usually occur in a power system, due to insulation failure, flashover, physical damage, or human errors. These faults may be either three phases in nature involving all three phases in a symmetrical manner, or asymmetrical, where usually only one or two phases may be involved. Faults may also be caused by either short circuits to earth or between live conductors, or may be caused by broken conductors (open circuit faults) in one or more phases. Sometimes simultaneous faults may occur involving both short circuit and broken conductor faults.

Balanced three-phase faults may be analyzed using an equivalent single-phase circuit. Symmetrical components are used to reduce the computational complexity with the study of asymmetrical three-phase faults.

18.4.1. Symmetrical component analysis

According to Fortscue’s theorem, unbalanced three-phase systems can be expressed in terms of three balanced components called symmetrical components, such as, positive sequence (balanced and having the same phase sequence as the unbalanced supply), negative sequence (balanced and having the opposite phase sequence to the unbalanced supply), and zero sequence (balanced but having the same phase and hence no phase sequence).

The phase components are the addition of symmetrical components, as shown in Figure 18.25 and are mathematically expressed as follows, such that it is possible to convert from either sequence components to phase components or vice versa.

$\begin{array}{l} a = a_{1} + a_{2} + a_{0} \\ b = b_{1} + b_{2} + b_{0} \\ c = c_{1} + c_{2} + c_{0} \end{array}$ $\begin{array}{l} a = a_{1} + a_{2} + a_{0} \\ b = b_{1} + b_{2} + b_{0} \\ c = c_{1} + c_{2} + c_{0} \end{array}$

Figure 18.25 Phase component in terms of symmetrical components.

Where a, b, and c are phase components and subscript 1 represents positive sequence, subscript 2 represents negative sequence and subscript 0 represents zero sequence components in an unbalanced systems.

18.4.2. Definition of the operator α

With the balanced components, the angle is 120°. Complex operator j is defined as

\sqrt{- 1}

$\sqrt{- 1}$

, that is, a vector of unit magnitude rotated anticlockwise by an angle of 90°.

$that is, j = \sqrt{- 1} = 1 ∠ 90 °$ $that is, j = \sqrt{- 1} = 1 ∠ 90 °$

Similarly a new complex operator α is defined with a magnitude of unity and when operated on any complex number rotates it anticlockwise by an angle of 120°:

$that is α = 1 ∠ 120 ° = - 0 .500 + j 0 .866$ $that is α = 1 ∠ 120 ° = - 0 .500 + j 0 .866$

Some properties of α:

$\begin{array}{l} α = 1 ∠ 120 ° in anticlockwise direction \\ α^{2} = 1 ∠ 120 ° or 1 ∠ - 120 ° \\ α^{3} = 1 ∠ 120 ° or 1 \\ that is, α^{3} - 1 = (α - 1) (α^{2} + α +1) = 0. \end{array}$ $\begin{array}{l} α = 1 ∠ 120 ° in anticlockwise direction \\ α^{2} = 1 ∠ 120 ° or 1 ∠ - 120 ° \\ α^{3} = 1 ∠ 120 ° or 1 \\ that is, α^{3} - 1 = (α - 1) (α^{2} + α +1) = 0. \end{array}$

Since α is complex, it cannot be equal to 1, so that α – 1 cannot be zero.

$∴ α^{2} + α + 1 = 0$ $∴ α^{2} + α + 1 = 0$

The sequence components of the unbalanced quantity, with each of the components written in terms of phase a components and the operator α, is given in Figure 18.26.

Figure 18.26 Expressing components in terms of phase a.

It is inferred that even in the unbalanced systems, phase b and phase c requires no measurements and can be computed with reference to the quantities measured in phase a itself. All the sequence components in terms of the quantities for the a phase using the properties of rotation can be expressed as,

$\begin{array}{l} a = a_{0} + a_{1} + a_{2} \\ b = a_{0} + α^{2} a_{1} + α a_{2} \\ c = a_{0} + α a_{1} + α^{2} a_{2} . \end{array}$ $\begin{array}{l} a = a_{0} + a_{1} + a_{2} \\ b = a_{0} + α^{2} a_{1} + α a_{2} \\ c = a_{0} + α a_{1} + α^{2} a_{2} . \end{array}$

This can be written in matrix form.

$\begin{array}{l} [\begin{array}{c} a \\ b \\ c \end{array}] = [\begin{array}{c} 1 & 1 & 1 \\ 1 & α^{2} & α \\ 1 & α & α^{2} \end{array}] [\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \end{array}] \\ \begin{array}{c} V_{a} = a_{11} V_{1} + a_{12} V_{2} + a_{13} V_{3} = V_{a_{1}} + V_{a_{2}} + V_{a_{0}} \\ V_{b} = a_{21} V_{1} + a_{22} V_{2} + a_{23} V_{3} = V_{b_{1}} + V_{b_{2}} + V_{b_{0}} \\ V_{c} = a_{31} V_{1} + a_{32} V_{2} + a_{33} V_{3} = V_{c_{1}} + V_{c_{2}} + V_{c_{0}} \end{array} \end{array}$ $\begin{array}{l} [\begin{array}{c} a \\ b \\ c \end{array}] = [\begin{array}{c} 1 & 1 & 1 \\ 1 & α^{2} & α \\ 1 & α & α^{2} \end{array}] [\begin{array}{c} a_{0} \\ a_{1} \\ a_{2} \end{array}] \\ \begin{array}{c} V_{a} = a_{11} V_{1} + a_{12} V_{2} + a_{13} V_{3} = V_{a_{1}} + V_{a_{2}} + V_{a_{0}} \\ V_{b} = a_{21} V_{1} + a_{22} V_{2} + a_{23} V_{3} = V_{b_{1}} + V_{b_{2}} + V_{b_{0}} \\ V_{c} = a_{31} V_{1} + a_{32} V_{2} + a_{33} V_{3} = V_{c_{1}} + V_{c_{2}} + V_{c_{0}} \end{array} \end{array}$

From Figure 18.26:

$\begin{array}{l} V_{b_{1}} = α^{2} V_{a_{1}} \\ V_{b_{2}} = α V_{a_{2}} \\ V_{c_{1}} = α V_{a_{1}} \\ V_{c_{2}} = α^{2} V_{a_{2}} \\ V_{b_{0}} = V_{c_{0}} = V_{a_{0}} \\ ∴ The phase components of an unbalanced system can be represented using the symmetrical components . \\ V_{a} = V_{a_{1}} + V_{a_{2}} + V_{a_{0}} \\ ∴ V_{b} = α^{2} V_{a_{1}} + α V_{a_{2}} + V_{a_{0}} \\ ∴ V_{c} = α V_{a_{1}} + α^{2} V_{a_{2}} + V_{a_{0}} . \end{array}$ $\begin{array}{l} V_{b_{1}} = α^{2} V_{a_{1}} \\ V_{b_{2}} = α V_{a_{2}} \\ V_{c_{1}} = α V_{a_{1}} \\ V_{c_{2}} = α^{2} V_{a_{2}} \\ V_{b_{0}} = V_{c_{0}} = V_{a_{0}} \\ ∴ The phase components of an unbalanced system can be represented using the symmetrical components . \\ V_{a} = V_{a_{1}} + V_{a_{2}} + V_{a_{0}} \\ ∴ V_{b} = α^{2} V_{a_{1}} + α V_{a_{2}} + V_{a_{0}} \\ ∴ V_{c} = α V_{a_{1}} + α^{2} V_{a_{2}} + V_{a_{0}} . \end{array}$

Similarly,

$\begin{array}{l} ∴ I_{a} = I_{a_{1}} + I_{a_{2}} + I_{a_{0}} \\ ∴ I_{b} = α^{2} I_{a_{1}} + α I_{a_{2}} + I_{a_{0}} \\ ∴ I_{c} = α I_{a_{1}} + α^{2} I_{a_{2}} + I_{a_{0}} . \end{array}$ $\begin{array}{l} ∴ I_{a} = I_{a_{1}} + I_{a_{2}} + I_{a_{0}} \\ ∴ I_{b} = α^{2} I_{a_{1}} + α I_{a_{2}} + I_{a_{0}} \\ ∴ I_{c} = α I_{a_{1}} + α^{2} I_{a_{2}} + I_{a_{0}} . \end{array}$

Problem 18.12

Given the phase voltages V_a, V_b, and V_c find the symmetrical components

V_{a_{0}}

$V_{a_{0}}$

V_{a_{1}}

$V_{a_{1}}$

, and

V_{a_{2}}

$V_{a_{2}}$

and find the average three phase power in terms of symmetrical components.

Solution

$\begin{array}{l} V_{a} + V_{b} + V_{c} = 3 V_{a_{0}} \\ ∴ V_{a_{0}} = \frac{1}{3} (V_{a} + V_{b} + V_{c}) . \end{array}$ $\begin{array}{l} V_{a} + V_{b} + V_{c} = 3 V_{a_{0}} \\ ∴ V_{a_{0}} = \frac{1}{3} (V_{a} + V_{b} + V_{c}) . \end{array}$

To get

V_{a_{1}}

$V_{a_{1}}$

, multiply by 1, α, and α²:

$\begin{array}{l} α^{2} V_{c} + α V_{b} + V_{a} = V_{a_{1}} + α V_{b} + α^{2} V_{c} = V_{a_{1}} (1 + α^{3} + α^{3}) + V_{a_{2}} (1 + α^{2} + α^{4}) + V_{a_{0}} (1 + α + α^{2}) = 3 V_{a_{1}} \\ ∴ V_{a_{1}} = \frac{1}{3} (V_{a} + α V_{b} + α^{2} V_{c}) . \end{array}$ $\begin{array}{l} α^{2} V_{c} + α V_{b} + V_{a} = V_{a_{1}} + α V_{b} + α^{2} V_{c} = V_{a_{1}} (1 + α^{3} + α^{3}) + V_{a_{2}} (1 + α^{2} + α^{4}) + V_{a_{0}} (1 + α + α^{2}) = 3 V_{a_{1}} \\ ∴ V_{a_{1}} = \frac{1}{3} (V_{a} + α V_{b} + α^{2} V_{c}) . \end{array}$

Similarly to get

V_{a_{2}}

$V_{a_{2}}$

, multiply by 1, α², and α,

$\begin{array}{l} α V_{c} + α^{2} V_{b} + V_{a} = 3 V_{a_{2}} \\ ∴ V_{a_{2}} = \frac{1}{3} (V_{a} + α^{2} V_{b} + α V_{c}) . \end{array}$ $\begin{array}{l} α V_{c} + α^{2} V_{b} + V_{a} = 3 V_{a_{2}} \\ ∴ V_{a_{2}} = \frac{1}{3} (V_{a} + α^{2} V_{b} + α V_{c}) . \end{array}$

Average three-phase power in terms of symmetrical components is,

$\begin{array}{l} P + j Q = V_{a} I_{a}^{*} + V_{b} I_{b}^{*} + V_{c} I_{c}^{*} = [\begin{array}{c} V_{a} & V_{b} & V_{c} \end{array}] {[\begin{array}{c} I_{a} \\ I_{b} \\ I_{c} \end{array}]}^{*} \\ [\begin{array}{c} V_{a} \\ V_{b} \\ V_{c} \end{array}] = [\begin{array}{c} 1 & 1 & 1 \\ 1 & α^{2} & α \\ 1 & α & α^{2} \end{array}] [\begin{array}{c} V_{a_{0}} \\ V_{a_{1}} \\ V_{a_{2}} \end{array}] = AV \\ {[\begin{array}{c} V_{a} \\ V_{b} \\ V_{c} \end{array}]}^{T} = {(AV)}^{T} = V^{T} A^{T} . \end{array}$ $\begin{array}{l} P + j Q = V_{a} I_{a}^{*} + V_{b} I_{b}^{*} + V_{c} I_{c}^{*} = [\begin{array}{c} V_{a} & V_{b} & V_{c} \end{array}] {[\begin{array}{c} I_{a} \\ I_{b} \\ I_{c} \end{array}]}^{*} \\ [\begin{array}{c} V_{a} \\ V_{b} \\ V_{c} \end{array}] = [\begin{array}{c} 1 & 1 & 1 \\ 1 & α^{2} & α \\ 1 & α & α^{2} \end{array}] [\begin{array}{c} V_{a_{0}} \\ V_{a_{1}} \\ V_{a_{2}} \end{array}] = AV \\ {[\begin{array}{c} V_{a} \\ V_{b} \\ V_{c} \end{array}]}^{T} = {(AV)}^{T} = V^{T} A^{T} . \end{array}$

But A^T = A as α and α² are conjugate.

$\begin{array}{l} {[\begin{array}{c} I_{a} \\ I_{b} \\ I_{c} \end{array}]}^{*} = {[[\begin{array}{c} 1 & 1 & 1 \\ 1 & α^{2} & α \\ 1 & α & α^{2} \end{array}] [\begin{array}{c} I_{a_{0}} \\ I_{a_{1}} \\ I_{a_{2}} \end{array}]]}^{*} \\ P + j Q = [\begin{array}{c} V_{a_{0}} & V_{a_{1}} & V_{a_{2}} \end{array}] [\begin{array}{c} 1 & 1 & 1 \\ 1 & α^{2} & α \\ 1 & α & α^{2} \end{array}] [\begin{array}{c} 1 & 1 & 1 \\ 1 & α & α^{2} \\ 1 & α^{2} & α \end{array}] [\begin{array}{c} I_{a_{0}} \\ I_{a_{1}} \\ I_{a_{2}} \end{array}] \\ P + j Q = 3 [\begin{array}{c} V_{a_{0}} I_{a_{0}}^{*} & V_{a_{1}} I_{a_{1}}^{*} & V_{a_{2}} I_{a_{2}}^{*} \end{array}] . \end{array}$ $\begin{array}{l} {[\begin{array}{c} I_{a} \\ I_{b} \\ I_{c} \end{array}]}^{*} = {[[\begin{array}{c} 1 & 1 & 1 \\ 1 & α^{2} & α \\ 1 & α & α^{2} \end{array}] [\begin{array}{c} I_{a_{0}} \\ I_{a_{1}} \\ I_{a_{2}} \end{array}]]}^{*} \\ P + j Q = [\begin{array}{c} V_{a_{0}} & V_{a_{1}} & V_{a_{2}} \end{array}] [\begin{array}{c} 1 & 1 & 1 \\ 1 & α^{2} & α \\ 1 & α & α^{2} \end{array}] [\begin{array}{c} 1 & 1 & 1 \\ 1 & α & α^{2} \\ 1 & α^{2} & α \end{array}] [\begin{array}{c} I_{a_{0}} \\ I_{a_{1}} \\ I_{a_{2}} \end{array}] \\ P + j Q = 3 [\begin{array}{c} V_{a_{0}} I_{a_{0}}^{*} & V_{a_{1}} I_{a_{1}}^{*} & V_{a_{2}} I_{a_{2}}^{*} \end{array}] . \end{array}$

18.4.3. Line-to-ground fault with Z_f [3]

Consider a three-phase system with a fault impedance of Z_f and the neutral impedance Z_n, as shown in Figure 18.27a.

Figure 18.27 (a) A three-phase unloaded alternator with neutral grounded through impedance Z_n and fault impedance Z_f, L–G fault and (b) interconnection of sequence network for L–G fault.

From Figure 18.27a,

$\begin{array}{c} V_{a} = I_{a} Z_{f} & I_{b} = 0 & I_{c} = 0 . \end{array}$ $\begin{array}{c} V_{a} = I_{a} Z_{f} & I_{b} = 0 & I_{c} = 0 . \end{array}$

The sequence network equations are,

$V_{a_{0}} = - I_{a_{0}} Z_{0} V_{a_{1}} = E_{a} - I_{a_{1}} Z_{1} V_{a_{2}} = - I_{a_{2}} Z_{2}$ $V_{a_{0}} = - I_{a_{0}} Z_{0} V_{a_{1}} = E_{a} - I_{a_{1}} Z_{1} V_{a_{2}} = - I_{a_{2}} Z_{2}$

Substituting I_b and I_c values, the solution of these equations gives the unknown quantities,

$\begin{array}{l} I_{a_{1}} = \frac{1}{3} (I_{a} + α I_{b} + α^{2} I_{c}) = \frac{1}{3} I_{a} \\ I_{a_{2}} = \frac{1}{3} (I_{a} + α^{2} I_{b} + α I_{c}) = \frac{1}{3} I_{a} \\ I_{a_{0}} = \frac{1}{3} (I_{a} + I_{b} + I_{c}) = \frac{1}{3} I_{a} \\ I_{a_{1}} = I_{a_{2}} = I_{a_{0}} = \frac{I_{a}}{3} \to I_{a} = 3 I_{a_{1}} . \end{array}$ $\begin{array}{l} I_{a_{1}} = \frac{1}{3} (I_{a} + α I_{b} + α^{2} I_{c}) = \frac{1}{3} I_{a} \\ I_{a_{2}} = \frac{1}{3} (I_{a} + α^{2} I_{b} + α I_{c}) = \frac{1}{3} I_{a} \\ I_{a_{0}} = \frac{1}{3} (I_{a} + I_{b} + I_{c}) = \frac{1}{3} I_{a} \\ I_{a_{1}} = I_{a_{2}} = I_{a_{0}} = \frac{I_{a}}{3} \to I_{a} = 3 I_{a_{1}} . \end{array}$

V_a = I_a Z_f → I_a = V_a/Z_f, written in terms of symmetrical components:

$\begin{array}{r} V_{a_{1}} + V_{a_{2}} + V_{a_{0}} = V_{a} = 3 I_{a_{1}} Z_{f} \\ E_{a} - I_{a_{1}} Z_{1} - I_{a_{2}} Z_{2} - I_{a_{0}} Z_{0} = 3 I_{a} Z_{f} \\ E_{a} = I_{a_{1}} [Z_{1} + Z_{2} + Z_{0} + 3 Z_{f}] \\ I_{a_{1}} = \frac{E_{a}}{Z_{1} + Z_{2} + Z_{0} + 3 Z_{f}} . \end{array}$ $\begin{array}{r} V_{a_{1}} + V_{a_{2}} + V_{a_{0}} = V_{a} = 3 I_{a_{1}} Z_{f} \\ E_{a} - I_{a_{1}} Z_{1} - I_{a_{2}} Z_{2} - I_{a_{0}} Z_{0} = 3 I_{a} Z_{f} \\ E_{a} = I_{a_{1}} [Z_{1} + Z_{2} + Z_{0} + 3 Z_{f}] \\ I_{a_{1}} = \frac{E_{a}}{Z_{1} + Z_{2} + Z_{0} + 3 Z_{f}} . \end{array}$

Since

I_{a_{1}}

$I_{a_{1}}$

I_{a_{2}}

$I_{a_{2}}$

, and

I_{a_{0}}

$I_{a_{0}}$

are known,

V_{a_{1}}

$V_{a_{1}}$

V_{a_{2}}

$V_{a_{2}}$

V_{a_{0}}

$V_{a_{0}}$

can be calculated from the sequence network equations. Thereafter the fault current I_a can be calculated. The sequence network interconnections are shown in Figure 18.27b.

18.4.4. Line-to-line fault with Z_f

From Figure 18.28a,

$I_{a} = 0 I_{b} + I_{c} = 0 \to I_{b} = - I_{c} V_{b} = V_{c} + I_{b} Z_{f}$ $I_{a} = 0 I_{b} + I_{c} = 0 \to I_{b} = - I_{c} V_{b} = V_{c} + I_{b} Z_{f}$

Figure 18.28 (a) A three-phase unloaded alternator with neutral grounded through impedance Z_n and fault impedance Z_f, and (b) interconnection of sequence network, fault impedance Z_f, L–L fault.

and the sequence network equations are,

$\begin{matrix} V_{a_{1}} = E_{a} - I_{a_{1}} Z_{1} V_{a_{2}} = - I_{a_{2}} Z_{2} V_{a_{0}} = - I_{a_{0}} Z_{0} \\ I_{a_{1}} = \frac{1}{3} (I_{a} + α I_{b} + α^{2} I_{c}) = I_{b} \\ I_{a_{2}} = \frac{1}{3} (I_{a} + α^{2} I_{b} + α I_{c}) = - I_{b} \\ I_{a_{0}} = \frac{1}{3} (I_{a} + I_{b} + I_{c}) = 0 \\ I_{a_{0}} = 0, I_{a_{2}} = - I_{a_{1}} \\ V_{b} = V_{b} + I_{b} Z_{f} \\ V_{a_{0}} + α^{2} V_{a_{1}} + α V_{a_{2}} = V_{a_{0}} + α V_{a_{1}} + α^{2} V_{a_{2}} + (I_{a_{0}} + α^{2} I_{a_{1}} + α I_{a_{2}}) Z_{f} \\ (α^{2} - α) V_{a_{1}} = (α^{2} - α) V_{a_{2}} + (α^{2} - α) I_{a_{1}} Z_{f} as I_{a_{0}} = 0 \to V_{a_{0}} = 0 and I_{a_{2}} = - I_{a_{1}} \\ V_{a_{1}} = V_{a_{2}} + I_{a_{1}} Z_{f} . \end{matrix}$ $\begin{matrix} V_{a_{1}} = E_{a} - I_{a_{1}} Z_{1} V_{a_{2}} = - I_{a_{2}} Z_{2} V_{a_{0}} = - I_{a_{0}} Z_{0} \\ I_{a_{1}} = \frac{1}{3} (I_{a} + α I_{b} + α^{2} I_{c}) = I_{b} \\ I_{a_{2}} = \frac{1}{3} (I_{a} + α^{2} I_{b} + α I_{c}) = - I_{b} \\ I_{a_{0}} = \frac{1}{3} (I_{a} + I_{b} + I_{c}) = 0 \\ I_{a_{0}} = 0, I_{a_{2}} = - I_{a_{1}} \\ V_{b} = V_{b} + I_{b} Z_{f} \\ V_{a_{0}} + α^{2} V_{a_{1}} + α V_{a_{2}} = V_{a_{0}} + α V_{a_{1}} + α^{2} V_{a_{2}} + (I_{a_{0}} + α^{2} I_{a_{1}} + α I_{a_{2}}) Z_{f} \\ (α^{2} - α) V_{a_{1}} = (α^{2} - α) V_{a_{2}} + (α^{2} - α) I_{a_{1}} Z_{f} as I_{a_{0}} = 0 \to V_{a_{0}} = 0 and I_{a_{2}} = - I_{a_{1}} \\ V_{a_{1}} = V_{a_{2}} + I_{a_{1}} Z_{f} . \end{matrix}$

Now substituting for

V_{a_{1}}

$V_{a_{1}}$

and

V_{a_{2}}

$V_{a_{2}}$

from the sequence network equations we get,

$\begin{array}{r} E_{a} - I_{a_{1}} Z_{1} = - I_{a_{2}} Z_{2} + I_{a_{1}} Z_{f} \\ E_{a} - I_{a_{1}} Z_{1} = I_{a_{1}} (Z_{2} + Z_{f}) \\ I_{a_{1}} = \frac{E_{a}}{Z_{1} + (Z_{2} + Z_{f})} . \end{array}$ $\begin{array}{r} E_{a} - I_{a_{1}} Z_{1} = - I_{a_{2}} Z_{2} + I_{a_{1}} Z_{f} \\ E_{a} - I_{a_{1}} Z_{1} = I_{a_{1}} (Z_{2} + Z_{f}) \\ I_{a_{1}} = \frac{E_{a}}{Z_{1} + (Z_{2} + Z_{f})} . \end{array}$

The interconnections of the sequence network, are shown in Figure 18.28b.

18.4.5. Double line-to-ground fault through Z_f

Figure 18.29 shows the double line-to-ground (DLG) fault through impedance Z_f. From Figure 18.29,

$I_{a} = 0 = I_{a_{1}} + I_{a_{2}} + I_{a_{0}} V_{b} = V_{c} = (I_{b} + I_{c}) Z_{f}$ $I_{a} = 0 = I_{a_{1}} + I_{a_{2}} + I_{a_{0}} V_{b} = V_{c} = (I_{b} + I_{c}) Z_{f}$

Figure 18.29 (a) L–L–G faultthrough a fault impedance Z_f and neutral impedance Z_n and (b) interconnection of sequence networks for (a).

and the sequence network equations are,

$\begin{array}{l} V_{a_{1}} = E_{a} - I_{a_{1}} Z_{1} V_{a_{2}} = - I_{a_{2}} Z_{2} V_{a_{0}} = - I_{a_{0}} Z_{0} \\ V_{b} = V_{c} \to V_{a_{0}} + α^{2} V_{a_{1}} + α V_{a_{2}} = V_{a_{0}} + α V_{a_{1}} + α^{2} V_{a_{2}} \to V_{a_{1}} = V_{a_{2}} \\ V_{b} = (I_{b} + I_{c}) Z_{f} \\ V_{a_{0}} + α^{2} V_{a_{1}} + α V_{a_{2}} = [I_{a_{0}} + α^{2} I_{a_{1}} + α I_{a_{2}} + I_{a_{0}} + α I_{a_{1}} + α^{2} I_{a_{2}}] Z_{f} \\ V_{a_{0}} + (α^{2} - α) V_{a_{1}} = 2 (I_{a_{0}} + (α^{2} + α) (I_{a_{1}} + I_{a_{2}}) Z_{f} \\ V_{a_{0}} - V_{a_{1}} = [2 I_{a_{0}} (α^{2} + α) (- I_{a_{0}})] Z_{f} \\ V_{a_{0}} - V_{a_{1}} = 3 I_{a_{0}} Z_{f} \\ - I_{a_{0}} Z_{0} - E_{a} - I_{a_{1}} Z_{1} = I_{a_{0}} Z_{f} \\ I_{a_{0}} = - \frac{E_{a} - I_{a_{1}} Z_{1}}{Z_{0} + 3 Z_{f}} . \end{array}$ $\begin{array}{l} V_{a_{1}} = E_{a} - I_{a_{1}} Z_{1} V_{a_{2}} = - I_{a_{2}} Z_{2} V_{a_{0}} = - I_{a_{0}} Z_{0} \\ V_{b} = V_{c} \to V_{a_{0}} + α^{2} V_{a_{1}} + α V_{a_{2}} = V_{a_{0}} + α V_{a_{1}} + α^{2} V_{a_{2}} \to V_{a_{1}} = V_{a_{2}} \\ V_{b} = (I_{b} + I_{c}) Z_{f} \\ V_{a_{0}} + α^{2} V_{a_{1}} + α V_{a_{2}} = [I_{a_{0}} + α^{2} I_{a_{1}} + α I_{a_{2}} + I_{a_{0}} + α I_{a_{1}} + α^{2} I_{a_{2}}] Z_{f} \\ V_{a_{0}} + (α^{2} - α) V_{a_{1}} = 2 (I_{a_{0}} + (α^{2} + α) (I_{a_{1}} + I_{a_{2}}) Z_{f} \\ V_{a_{0}} - V_{a_{1}} = [2 I_{a_{0}} (α^{2} + α) (- I_{a_{0}})] Z_{f} \\ V_{a_{0}} - V_{a_{1}} = 3 I_{a_{0}} Z_{f} \\ - I_{a_{0}} Z_{0} - E_{a} - I_{a_{1}} Z_{1} = I_{a_{0}} Z_{f} \\ I_{a_{0}} = - \frac{E_{a} - I_{a_{1}} Z_{1}}{Z_{0} + 3 Z_{f}} . \end{array}$

Similarly making use of the relation

V_{a_{1}}

$V_{a_{1}}$

V_{a_{2}}

$V_{a_{2}}$

, expressing

I_{a_{2}}

$I_{a_{2}}$

in terms of

I_{a_{1}}

$I_{a_{1}}$

$\begin{matrix} E_{a} - I_{a_{1}} Z_{1} = - I_{a_{2}} Z_{2} \\ I_{a_{2}} = - \frac{E_{a} - I_{a_{1}} Z_{1}}{Z_{2}} . \end{matrix}$ $\begin{matrix} E_{a} - I_{a_{1}} Z_{1} = - I_{a_{2}} Z_{2} \\ I_{a_{2}} = - \frac{E_{a} - I_{a_{1}} Z_{1}}{Z_{2}} . \end{matrix}$

Now substituting the values of

I_{a_{2}}

$I_{a_{2}}$

and

I_{a_{0}}

$I_{a_{0}}$

in the equation,

$\begin{matrix} I_{a} = I_{a_{1}} + I_{a_{2}} + I_{a_{0}} = 0 \\ I_{a_{1}} = - \frac{E_{a} - I_{a_{1}} Z_{1}}{Z_{2}} - \frac{E_{a} - I_{a_{1}} Z_{1}}{Z_{0} + 3 Z_{f}} = 0 \\ I_{a_{1}} = \frac{E_{a}}{Z_{1} + (Z_{2} (Z_{0} + 3 Z_{f}) / Z_{2} + Z_{0} + 3 Z_{f})} . \end{matrix}$ $\begin{matrix} I_{a} = I_{a_{1}} + I_{a_{2}} + I_{a_{0}} = 0 \\ I_{a_{1}} = - \frac{E_{a} - I_{a_{1}} Z_{1}}{Z_{2}} - \frac{E_{a} - I_{a_{1}} Z_{1}}{Z_{0} + 3 Z_{f}} = 0 \\ I_{a_{1}} = \frac{E_{a}}{Z_{1} + (Z_{2} (Z_{0} + 3 Z_{f}) / Z_{2} + Z_{0} + 3 Z_{f})} . \end{matrix}$

The interconnection of the sequence network is shown in Figure 13.29b. Based on the above fault analysis discussions, it is inferred that

1. Normally if the system is balanced, the analysis is done in any one of the phase, which can be extended for the rest of the phases. Similarly, even if the system is unbalanced, the fault current can be calculated in phase a and can be extended for all other phases.

2. With the presence of circulating sequence currents, it is possible to discriminate between fault types. For example, it is proved that positive sequence currents flow irrespective of the fault types. Negative sequence currents flow only if the fault involves more than one phase. Zero sequence currents flow if the fault involves ground.

18.4.6. Sequence networks

In all respect, the positive sequence network is identical to the usual networks considered. Each synchronous machine must be considered as a source of EMF, which may vary in magnitude and phase position depending upon the distribution of power and reactive volt amperes, just prior to the occurrence of the fault. The positive sequence voltage at the point of fault will drop, the amount being dependent upon the type of faults; for three-phase faults it will be zero; for DLG fault, line-to-line fault, and single line-to-ground fault, it will be higher in the order stated.

The negative sequence network is in general quite similar to the positive sequence network except that since no negative sequence voltages are generated, the source of EMF is absent.

The zero sequence networks will be free of internal voltages, flow of current resulting from the voltage at the point of fault. The impedances to zero sequence current are very frequently different from positive or negative sequence currents. Transformer and generator impedances depend upon the type of star or delta connections.

The zero sequence equivalent circuits of three-phase transformers require special attention because of the possibility of various combinations. A general circuit for any combination is given in Figure 18.30a. Z₀ is the zero sequence impedance of the windings of the transformer. These are two series and two shunt switches. One series and one shunt switch are for both the sides separately. The series switch of a particular side is closed if it is star grounded and the shunt switch is closed if that side is delta connected, otherwise they are left open. For example, consider the transformer ∆/Y is connected with star ungrounded, as shown in Figure 18.30b. The switching arrangement shown in Figure 18.30c, since the primary is delta connected, the shunt switch of the primary side is closed and series is left open. The secondary is star ungrounded, therefore the series switch is left open and the shunt switch is also left open. The zero sequence network is shown in Figure 18.30d.